Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sampling With The Normal Distribution

  1. Feb 9, 2010 #1
    I have been puzzling over this problem for about a week now and cannot find the answer. In my opinion it is very theoretical, but I know I ma not the best mathematician on here so maybe someone else could look at this.

    Thank you.
  2. jcsd
  3. Feb 9, 2010 #2
    You need to derive the probability distribution for [tex]\bar{X}[/tex]. This is possible, since it is a sum of n quantities for which you know the distribution. Tthe distribution for [tex]\bar{X}[/tex] willl depend on n. You can then calculate the probability that [tex]|\bar{X}-\mu|>\sigma/2[/tex] in the ordinary way. This gives you an inequality for n. Then find the smallest n that satisfies the inequality.

    When you know n, you have completely specified the distribution for [tex]\bar{X}[/tex], and you can do part (b).

    So the first step is to find the distribution of a sum of quantities, expressed in terms of their individual distributions.

  4. Feb 9, 2010 #3
    Do you mean [tex]\bar{X}[/tex]~N([tex]\mu[/tex],[tex]\sigma[/tex]2/n)

    Where do I go from here if I knew the probability then I think I could manage.
  5. Feb 9, 2010 #4
    Cool, that's the same thing I got. From that you can write the explicit gaussian probability density for e.g. [tex]u:=\bar{X}-\mu[/tex]. Call it e.g. [tex]\rho_n(u)[/tex]. It will depend on n of course. It will simply the usual normalied gaussian centered around u=0, with variance [tex]\sigma^2/n[/tex] I think.

    Then the probability [tex]P(n)[/tex] that [tex]|u|>\sigma/2[/tex] is given as a integral that you can solve using the error function:

    P(n) = \int_{-\infty}^{-\sigma/2} \rho_n(u)du + \int^{\infty}_{\sigma/2} \rho_n(u)du

    Then find the smallest n such that P(n)>0.05. Then fix this value of n, and just do another simple integral to get the answer to part b.


  6. Feb 9, 2010 #5
    Got it now. It comes out n>16 so at least n=16. I used a different method which I will post now.
  7. Feb 9, 2010 #6


    =[tex]\phi[/tex]([tex]\sqrt{n}[/tex]/2) - (1-([tex]\phi[/tex]([tex]\sqrt{n}[/tex]/2))>0.95








    But n is integer
    Therefore n>16

    The minimum value of n is therefore 16
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook